Find the sum of the multiples of 6 from 200 to 1100
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First multiple after 200 = 204
Last multiple before 1100 = 1098
Here first term a = 204
Common Difference d = 6
Last term an = 1098
an = a + ( n - 1 )* d
=> 1098 = 204 + ( n - 1 ) * 6
=> 1098 = 204 + 6n - 6
=> 1098 = 198 + 6n
=> 1098 - 198 = 6n
=> 900 = 6n
=> 900 / 6 = n
=> n = 150
We know that the last sum of nth term is
Sn = n / 2 [ a + l ]
=> S150 = 150 / 2 [ 204 + 1098 ]
=> S150 = 75 × 1302
=> S150 = 97650.
Therefore, the sum of the multiples of 6 from 200 to 1100 is 97650.
Hope it helps!
First multiple after 200 = 204
Last multiple before 1100 = 1098
Here first term a = 204
Common Difference d = 6
Last term an = 1098
an = a + ( n - 1 )* d
=> 1098 = 204 + ( n - 1 ) * 6
=> 1098 = 204 + 6n - 6
=> 1098 = 198 + 6n
=> 1098 - 198 = 6n
=> 900 = 6n
=> 900 / 6 = n
=> n = 150
We know that the last sum of nth term is
Sn = n / 2 [ a + l ]
=> S150 = 150 / 2 [ 204 + 1098 ]
=> S150 = 75 × 1302
=> S150 = 97650.
Therefore, the sum of the multiples of 6 from 200 to 1100 is 97650.
Hope it helps!
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